Math 361, Spring 2014, Assignment 1
Carefully define the following terms, then give one example and one non-example of each:[edit]
- Homomorphism (of rings).
- Kernel (of a ring homomorphism).
- Ideal.
- Quotient ring.
Carefully state the following theorems (you need not prove them):[edit]
- Fundamental theorem on ring homomorphisms.
Solve the following problems:[edit]
- Section 26, problems 4, 12, 13, 14, 15, and 17.
Questions:[edit]
Solutions:[edit]
Definitions:[edit]
- Homomorphism (of rings).
Definition: A map $\phi$ of a ring $R$ into a ring $R'$ is a homomorphism if it satisfies:
$$\phi(a+b)=\phi(a)+\phi(b)$$ $$\phi(ab)=\phi(a)\phi(b)$$ for all $a,b\in R$Example:
The function $\phi :\mathbb{Z}\rightarrow\mathbb{Z}_n$ defined by $\phi(z) = z\enspace mod\enspace n$ is a ring homomorphism
Non-Example:
There is no homomorphism from $\phi :\mathbb{Z}_n\rightarrow\mathbb{Z}$ when $n>1$.
- Kernel (of a ring homomorphism).
Let a map $\phi : R\rightarrow R$ be a homomorphism of rings. The subring
$$\phi^{-1}[0'] = \{r\in R|\phi(r) = 0'\}$$is the kernel of $\phi$, denoted $Ker(\phi)$.
Example:
The function $\phi :\mathbb{Z}\rightarrow\mathbb{Z}_n$ defined by $\phi(z) = z\enspace mod\enspace n$ is a ring homomorphism with kernel $n\mathbb{Z}$
Non-Example:
Since there is no homomorphism from $\phi :\mathbb{Z}_n\rightarrow\mathbb{Z}$ when $n>1$, there is no kernel. - Ideal
Given a ring \(R\), an ideal of \(R\) is a subring \(I\) of \(R\), with the additional property that \(I\) absorbs left and right multiplication. That is to say, \(\forall a \in R: \forall i \in I: ai\in I, ia\in I\).
Example
The set \(\langle 5\rangle\) is an ideal of \(\mathbb{Z}\).
Non-example
- Quotient Ring
Given a ring \(R\) and an ideal \(I\) of \(R\), define a quotient ring \(R/I\) as the set of equivalence classes \(a+I\), where \(a+I ~ b+I\) if \(a-b \in I\). Define \((a+I)+(b+I) = (a+b)+I\) and \((a+I)(b+I)=(ab+I)\). This structure is a ring.
Example
\(\mathbb{Z}_7\) is quotient ring. (Ideals are cosets of the additive group.)
Non-example
Theorems:[edit]
- Fundamental Theorem on Ring Homomorphisms
Let \(R\) and \(S\) be rings, and \(\phi:R\rightarrow S\) a ring homomorphism. Then \(Im(\phi)\) is a ring, and there is an isomorphism from \(R/ker(\phi)\) to \(Im(\phi)\).